geometry: old and new - usna · conic: =) [125][136][246][345] = +[126][135][245][346] ... and...
TRANSCRIPT
Geometry: Old and New
Will Traves
Department of MathematicsUnited States Naval Academy∗
Roanoke CollegeMAA MD/DC/VA Section Meeting
25 APR 2015
* Any views or opinions presented in this talk are solely those of the presenter
and do not necessarily represent those of the U.S. Government
Traves MAA MD/DC/VA 25 APR 2015
Ancient Geometry
Line Arrangement due to Pappus of Alexandria (Synagogue; c. 340)
Richter-Gebert: 9 proofs in Perspectives on Projective Geometry, 2011
Traves MAA MD/DC/VA 25 APR 2015
Pascal’s Mystic Hexagon Theorem
Pascal: placed the 6 intersection points on a conic (1639)
Braikenridge and MacLaurin: proved the converse
Traves MAA MD/DC/VA 25 APR 2015
Why Mystic?
Traves MAA MD/DC/VA 25 APR 2015
The Projective Plane
P2 is a compactification of R2
P2 = R2 ∪ line at∞
Parallel lines meet at infinity - one point at∞ for each slope.Line at infinity wraps twice around R2.
Traves MAA MD/DC/VA 25 APR 2015
The Projective Plane: Thicken Line at Infinity
P2 is a compactification of R2
P2 = R2 ∪ line at∞
Parallel lines meet at infinity - one point at∞ for each slope.Line at infinity wraps twice around R2.
Traves MAA MD/DC/VA 25 APR 2015
Topology of the Projective Plane
Thicken line at infinity:P2 = disk ∪ Mobius band
P2 can’t be embedded in R3
(Conway, Gordon, Sachs (1983):linked triangles in K6)
! 20 !
Projective space is also the union of a disc in R2 and a Möbius strip, and is equivalent to the sphere S2 with a blow up at one point.
The Relationship between RP2 and R3.
Earlier we claimed that RP2 is not a subset of R3, that it does not “fit” into R3. For this proof we call upon Conway, Gordon and Sachs’ 1983 result (ams.org) that 6 points all cannot be linked to one another such that the total linking number of all triangles formed is even, in R3. We will give an example of 6 points linked in RP2 , K6, such that the linking number of the set is even. First, to demonstrate Conway, Gordon and Sachs’ proof, an example of K6, 6 points linked in R3:
Figure 16
The linking numbers of triangles 124 and 356 is 0 because they are not linked; they could be pulled apart from each other without being caught like a chain link.
The linking numbers of triangles 246 and 135 is 1 because they are linked together.
If you add all of the linking numbers of all sets of triangles in this particular linking, you will
find the sum to be 1 (246 and 135 are the only linked triangles). This is consistent with Conway, Gordon and Sachs who claimed that as long as we are working in R3 then we will have an odd total linking number.
Traves MAA MD/DC/VA 25 APR 2015
K6 embedded in P2
No linked trianglesTraves MAA MD/DC/VA 25 APR 2015
Bezout’s Theorem
Compactness of P2 allows us to count solutions:
Theorem (Bezout)Any two curves, without common components, defined by thevanishing of polynomials of degrees d1 and d2 meet in d1d2 points inP2, suitably interpreted.Lines meeting an Ellipse
y=0
4x2+9y2=36
y=1 y=2 y=3 Line meets curve in two
points (possibly imaginary).
Double point: tangency when y=2 4x2 + 9(4) = 36 so 4x2 = 0
It seems that such a curve of degree 1 always meets such a curve of degree 2 in (2)(1) points, if we count them properly.
Traves MAA MD/DC/VA 25 APR 2015
Projective Coordinates: Mobius’s model for P2
Mobius: P2 = {lines through the origin}Line through (x , y , z) has coordinates (x : y : z) with
(x : y : z) ∼ (λx : λy : λz) for λ 6= 0
Lines meeting z = 1 are of form (x : y : 1) ∼ (x , y)Other lines (in xy -plane) form points at infinity (x : y : 0)
Traves MAA MD/DC/VA 25 APR 2015
Lines in P2: I
Lines in P2 correspond to planes through originLine {(x : 0 : z)} meets line {(0 : y : z)} at point (0:0:1).
Traves MAA MD/DC/VA 25 APR 2015
Lines in P2: II
Parallel lines: {(x : 0 : z)} meets {(x : z : z)} at point (1:0:0).
Traves MAA MD/DC/VA 25 APR 2015
Homogenization
Can’t talk about the parabola y = x2:
If we scale the coordinates (x : y : z) = (λx : λy : λz) then we’drequire
λy = λ2x2
for all λ (true only for the origin (x , y) = (0,0)).
Curves in P2 are defined by the vanishing of homogeneouspolynomials:
y − x2 = 0←→ yz − x2 = 0
Traves MAA MD/DC/VA 25 APR 2015
Bashelor’s Work
Enumerative Geometry of Conics: How many conics pass through ppoints and are tangent to ` lines and c conics in general position?USNA Trident Project
This talk reports on joint work with MIDN Andrew Bashelor and my colleague at USNA, Amy Ksir.
Our work grew out of Bashelor’s Trident project, a full-year undergraduate research project focused on enumerative algebraic geometry.
Traves MAA MD/DC/VA 25 APR 2015
Conics through 5 points
ax2 + bxy + cy2 + dx + ey + f = 0ax2 + bxy + cy2 + dxz + eyz + fz2 = 0
(a : b : c : d : e : f ) ∈ R6/ ∼= P5
Point conditions force (a : b : c : d : e : f ) to lie on a hyperplane.
(x0 : y0 : z0) on C ⇐⇒ ax20 + bx0y0 + cy2
0 + dx0z0 + ey0z0 + fz20 = 0.
Intersecting 5 hyperplanes in P5 gives a single point corresponding tothe one conic through all 5 points.
Traves MAA MD/DC/VA 25 APR 2015
Determinants
Given three points in P2 we can list them as columns of a 3x3 matrix, ax bx cxay by cyaz bz cz
.
The determinant [abc] of this matrix measures six times the volume ofthe tetrahedron with edges a, b and c.
[abc] = 0 ⇐⇒ points a,b, c are collinear.
Can recover coordinates from full knowledge of determinants.
Traves MAA MD/DC/VA 25 APR 2015
Plucker Relations
The determinants of the 3x3 submatrices of a larger matrix satisfyquadratic relations.
For example, given five points in P2 we form the matrix ax bx cx dx exay by cy dy eyaz bz cz dz ez
.Cramer’s Theorem implies that
[abc][ade]− [abd ][ace] + [abe][acd ] = 0.
In particular, if [abc] = 0 then [abe][acd ] = [abd ][ace].
Traves MAA MD/DC/VA 25 APR 2015
Conics and Pascal’s Theorem
Six points lie on a conic⇐⇒ [abc][aef ][dbf ][dec]= [def ][dbc][aec][abf ].
conic: =⇒ [125][136][246][345] = +[126][135][245][346][159] = 0 =⇒ [157][259] = −[125][597][168] = 0 =⇒ [126][368] = +[136][268][249] = 0 =⇒ [245][297] = −[247][259][267] = 0 =⇒ [247][268] = −[246][287][348] = 0 =⇒ [346][358] = +[345][368][357] = 0 =⇒ [135][587] = −[157][358][257][987] = 0 ⇐= [297][587] = +[287][597]
If points 2, 5, and 7 are not collinear, [257] 6= 0 so [987] = 0.
Traves MAA MD/DC/VA 25 APR 2015
Computer Assistance I: Conjecture + Proof
Conjecture (Kepler)
No packing of spheres covers more than π/3√
2 (approximately 74%)of the filled volume.
History: Harriot and Sir Walter Raleigh (1591) and Kepler (1611)Gauss (regular lattices; 1631) and Fejes Toth (1953)Tom Hales and Sam Ferguson (1992-2006)FlysPecK (Formal Proof of Kepler; 2014)
Traves MAA MD/DC/VA 25 APR 2015
Computer Assistance II
Prediction: Computers will become our mathematical assistants, vastlyraising the level of our mathematical reasoning (c.f. advanced chessand CAD/origami).
Traves MAA MD/DC/VA 25 APR 2015
Computer Assistance III
We already have strong computer and robotic assistance:
Mind controlled Deka-arm
Traves MAA MD/DC/VA 25 APR 2015
Computer Assistance IV
Wild Conjecture: In the (perhaps distant) future the division betweenhuman and computer will become less and less distinct.
Time Magazine 2011
Traves MAA MD/DC/VA 25 APR 2015
My current work
There is at least one degree 3 curve through every set of 9 points.
QuestionWhen do 10 points lie on a plane curve of degree 3?
Smooth curves of degree 3 are called elliptic curves and play a role inboth elliptic curve cryptography and in Wiles’s proof of Fermat’sLast Theorem.
The 10 points lie on a cubic when the determinant of a 10× 10 matrixis zero (25 million terms).
Reiss (1842) wrote out a 20 term polynomial of degree 10 in the 3× 3brackets that computes the determinant faster.
Traves MAA MD/DC/VA 25 APR 2015
Ruler and Compass Construction
Construction (T and Wehlau)We developed a straightedge-and-compass construction that checkswhether 10 points lie on a cubic.
Such constructive questions are now back in vogue since theyleverage all sorts of ideas in computational geometry.
Traves MAA MD/DC/VA 25 APR 2015
The Key Idea: Cayley-Bacharach
10 points on a cubic precisely when 6 auxiliary points on a conic.
We construct the 6 points using straightedge and compass and theninvoke Pascal’s Theorem.
Traves MAA MD/DC/VA 25 APR 2015
Straightedge Construction
Meet and Join Algebra: Allows algebraic formulation ofstraightedge-only constructions
Theorem (after Sturmfels and Whiteley)There exists a straightedge construction to determine if 10 points lie ona cubic (with about 100 million lines).
Traves MAA MD/DC/VA 25 APR 2015